UVALive 6886 Golf Bot

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn = 100005;
const double pi = acos(-1.0);
const int low(int x){ return x&-x; }//树状数组的lowbit，可以计算二进制最右边的1
int n, m;

class FFT
{
private:
  const static int maxn = 560000;//要注意长度是2^k方
  class Plural
  {
  public:
    double x, y;
    Plural(double x = 0.0, double y = 0.0) :x(x), y(y){}
    Plural operator +(const Plural &a)
    {
      return Plural(x + a.x, y + a.y);
    }
    Plural operator -(const Plural &a)
    {
      return Plural(x - a.x, y - a.y);
    }
    Plural operator *(const Plural &a)
    {
      return Plural(x*a.x - y*a.y, x*a.y + y*a.x);
    }
    Plural operator /(const double &u)
    {
      return Plural(x / u, y / u);
    }
  };//定义复数的相关运算
  Plural x[maxn], x1[maxn], x2[maxn];
  Plural y[maxn], y1[maxn], y2[maxn];
  int ans[maxn], X[maxn];
  int n, len;
public:
  int reverse(int x)
  {
    int ans = 0;
    for (int i = 1, j = n >> 1; j; i <<= 1, j >>= 1) if (x&i) ans |= j;
    return ans;
  }//数字倒序,FFT的初始步骤
  Plural w(double x, double y)
  {
    return Plural(cos(2 * pi * x / y), -sin(2 * pi * x / y));
  }
  bool setx()
  {
    if (scanf("%d", &len) == EOF) return false;
    for (n = 400000; n != low(n); n += low(n));
    for (int i = 0; i < n; i++) { x[i] = Plural(0, 0); ans[i] = 0; }
    for (int i = n = 0; i < len; i++)
    {
      scanf("%d", &X[i]);
      x[X[i]].x += 1.0;
      n = max(n, X[i]);
    }
    for (n = n + n - 1; n != low(n); n += low(n));
    return true;
  }
  void fft(Plural*x, Plural*y, int flag)
  {
    for (int i = 0; i < n; i++) y[i] = x[reverse(i)];
    for (int i = 1; i < n; i <<= 1)
    {
      Plural uu = w(flag, i + i);
      for (int j = 0; j < n; j += i + i)
      {
        Plural u(1, 0);
        for (int k = j; k < j + i; k++)
        {
          Plural a = y[k];
          //w(flag*(k - j), i + i) 可以去掉u和uu用这个代替，精度高些，代价是耗时多了
          Plural b = u * y[k + i];
          y[k] = a + b;
          y[k + i] = a - b;
          u = u*uu;
        }
      }
    }
    if (flag == -1) for (int i = 0; i < n; i++) y[i] = y[i] / n;
  }//1是FFT，-1是IFFT，答案数组是y数组
  void solve()
  {
    fft(x, y, 1);
    for (int i = 0; i < n; i++) y[i] = y[i] * y[i];
    fft(y, x, -1);
    for (int i = 0; i < n; i++) ans[i] = (int)(x[i].x + 0.5);//调整精度
    for (int i = 0; i < len; i++) ans[X[i]]++;
    int u, v = 0;
    for (scanf("%d", &len); len; len--)
    {
      scanf("%d", &u);
      if (ans[u]) v++;
    }
    printf("%d", v);
    putchar(10);
  }
}fft;

int main()
{
  while (fft.setx())
  {
    fft.solve();
  }
  return 0;
}